gdb ouvre des files descriptor lorsqu'il est démarré.
Il devient donc très simple de détecter gdb depuis un binaire:
mitsurugi@mitsu:~/gdb$ cat close3.c
#include <stdio.h>
#include <stdlib.h>
cint main(void) {
if (close(3)==0) {
printf("Debugger detected, Bye\n");
exit(0);
}
printf("Je suis dans le main\n");
}
mitsurugi@mitsu:~/gdb$
et c'est tout:
mitsurugi@mitsu:~/gdb$ gdb -n -q ./close3
Reading symbols from /home/mitsurugi/gdb/close3...done.
(gdb) r
Starting program: /home/mitsurugi/gdb/./close3
Debugger detected, Bye
[Inferior 1 (process 15890) exited normally]
(gdb) q
mitsurugi@mitsu:~/gdb$ ./close3
Je suis dans le main
mitsurugi@mitsu:~/gdb$
Simple et efficace.
Ca se contourne tout aussi facilement, il existe plein de doc sur le sujet.
Le problème avec cette méthode, c'est que n'importe quel reverser la voit immédiatement. On pourrait imaginer des crackmes plus fins avec une détection silencieuse du genre:
mitsurugi@mitsu:~/gdb$ cat close3.c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char etat[32] = "gentil";
int main(void) {
if (close(3)==0) {
strcpy(etat,"mechant");
}
printf("Je suis dans le main\n");
printf("Je suis %s.\n", etat);
}
Ce qui donne:
mitsurugi@mitsu:~/gdb$ ./close3
Je suis dans le main
Je suis gentil.
mitsurugi@mitsu:~/gdb$ gdb -n -q ./close3
Reading symbols from /home/mitsurugi/gdb/close3...done.
(gdb) r
Starting program: /home/mitsurugi/gdb/./close3
Je suis dans le main
Je suis mechant.
[Inferior 1 (process 16571) exited with code 021]
(gdb) q
mitsurugi@mitsu:~/gdb$
Cet exemple n'est pas très représentatif, mais on peut imaginer un changement de magic number, un appel à une fonction qui ne se fait pas ou ce genre de choses.
mardi 10 décembre 2013
jeudi 5 décembre 2013
linux anti debug trick
Sous linux, tous les outils de décompilation s'attendent à avoir des binaires bien formés.
Prenons l'exemple du binaire crackme.01.32 de crackmes.de. L'antidebug est simple à réaliser et très efficace comme nous allons voir: la partie section header du binaire est erronée. Un hexdump le montre bien:
mitsurugi@mitsu:~/crackmes$ hexdump -C crackme.01.32 | head -4
00000000 7f 45 4c 46 01 01 01 00 00 00 00 00 00 00 00 00 |.ELF............|
00000010 02 00 03 00 ff ff ff ff b0 84 04 08 34 00 00 00 |............4...|
00000020 ff ff ff ff 00 00 00 00 ff ff 20 00 08 00 ff ff |.......... .....|
00000030 ff ff ff ff 06 00 00 00 34 00 00 00 34 80 04 08 |........4...4...|
mitsurugi@mitsu:~/crackmes$
Les sections headers sont remplacées par des ff ff, voir le post précédent avec le grand tableau de l'elf101.
Et c'est suffisant pour mettre hors jeu les debuggers classiques sous linux que j'ai pu essayer:
mitsurugi@mitsu:~/crackmes$ objdump -d crackme.01.32
objdump: crackme.01.32: File format not recognized
mitsurugi@mitsu:~/crackmes$ gdb -q crackme.01.32
"/home/mitsurugi/crackmes/crackme.01.32": not in executable format: Format de fichier non reconnu
(gdb) q
mitsurugi@mitsu:~/crackmes$ readelf -s crackme.01.32
readelf: Error: Unable to seek to 0xffffffff for section headers
readelf: Error: Unable to seek to 0xffffffff for section headers
mitsurugi@mitsu:~/crackmes$
Les debuggers spécialisés chargent le binaire, mais semblent avoir des problèmes par la suite. edb a crashé assez vite, radare2 charge le binaire, mais ne fait aucun mapping par la suite: tous octets du binaires sont simplement plaqué linéairement à partir de l'adresse 0x08048000. J'ai ouvert un bug chez radare2 du coup.
En fait, seul IDA semble bien s'en sortir. Le post de blog http://em386.blogspot.de/2006/10/elf-no-section-header-no-problem.html qui date de 2006 arrive exactement à la même conclusion.
Je trouve quand même ça fou que tous les debuggers s’appuient sur des sections qui ne sont pas utilisés à l'exécution. Cela signifie qu'il doit être possible de duper ces debuggers/reversers en forgeant des en-têtes de sections bien choisis. Comme le dit le post de blog plus haut, gdb est un peu dédié à l'analyse de code dont on a les sources, donc c'est normal qu'il ne soit pas robuste aux binaires mal fichus, mais c'est quand même dommage qu'il ne sache pas s'en sortir sans section headers.
J'ai envoyé une solution à crackmes.de en anglais en donnant une méthode sans IDA pour trouver la solution. Ca m'a permis de bien disséquer de l'ELF, mais c'est relativement long.
Solution:
Launch the binary to begin:
mitsurugi@mitsu:~/crackmes$ ./crackme.01.32
Please tell me my password: password
No! No! No! No! Try again.
mitsurugi@mitsu:~/crackmes$ gdb -q ./crackme.01.32
"/home/mitsurugi/crackmes/./crackme.01.32": not in executable format: File format not recognized
gdb$ q
mitsurugi@mitsu:~/crackmes$
At a first glance we can see an antidebug trick by wiping the section header:
mitsurugi@mitsu:~/crackmes$ readelf -s crackme.01.32
readelf: Error: Unable to seek to 0xffffffff for section headers
readelf: Error: Unable to seek to 0xffffffff for section headers
mitsurugi@mitsu:~/crackmes$
objdump fails at analyzing it, and gdb too. If we strings the binary we can also see:
mitsurugi@mitsu:~/crackmes$ strings -n10 crackme.01.32
/lib/ld-linux.so.2
_IO_stdin_used
__libc_start_main
__gmon_start__
I'm sorry GDB! You are not allowed!
Tracing is not allowed... Bye
Please tell me my password:
The password is correct!
Congratulations!!!
No! No! No! No! Try again.
GCC: (GNU) 4.8.0
.note.ABI-tag
.note.gnu.build-id
.gnu.version
.gnu.version_r
.eh_frame_hdr
.init_array
.fini_array
mitsurugi@mitsu:~/crackmes$
The message about GDB let us assume that there are some other anti-gdb tricks, ptrace maybe:
mitsurugi@mitsu:~/crackmes$ strings crackme.01.32 | grep ptrace
ptrace
mitsurugi@mitsu:~/crackmes$
So it's time to disass it.We know that section headers are not usefull at all when running a binary, but program header are mandatory. Let's see what we have:
mitsurugi@mitsu:~/crackmes$ readelf -l crackme.01.32
readelf: Error: Unable to seek to 0xffffffff for section headers
readelf: Error: Unable to seek to 0xffffffff for section headers
Elf file type is EXEC (Executable file)
Entry point 0x80484b0
There are 8 program headers, starting at offset 52
Program Headers:
Type Offset VirtAddr PhysAddr FileSiz MemSiz Flg Align
PHDR 0x000034 0x08048034 0x08048034 0x00100 0x00100 R E 0x4
INTERP 0x000134 0x08048134 0x08048134 0x00013 0x00013 R 0x1
[Requesting program interpreter: /lib/ld-linux.so.2]
LOAD 0x000000 0x08048000 0x08048000 0x00a28 0x00a28 R E 0x1000
LOAD 0x000a28 0x08049a28 0x08049a28 0x00140 0x00180 RW 0x1000
DYNAMIC 0x000a38 0x08049a38 0x08049a38 0x000e8 0x000e8 RW 0x4
NOTE 0x000148 0x08048148 0x08048148 0x00044 0x00044 R 0x4
GNU_EH_FRAME 0x0008d4 0x080488d4 0x080488d4 0x00044 0x00044 R 0x4
GNU_STACK 0x000000 0x00000000 0x00000000 0x00000 0x00000 RW 0x4
mitsurugi@mitsu:~/crackmes$
So we can map some memory addresses with file offset.
First LOAD section:
From 0x08048000 to 0x08048a28, it's a linear mapping with the file from offset 0. But for bytes next, we have to make a gap of 0x1000.
Second LOAD section:
From 0x08049a28 to 0x08049a28+0x180 we have to map offset from files 0xa2 to 0xa28+0x140..
The DYNAMIC section, NOTE, GNU_EH_FRAME are not used for solving this crackme.
And now, with the help of ndisasm, we can solve the challenge by starting at entry point:
mitsurugi@mitsu:~/crackmes$ ndisasm -au -o 0x08048000 -e 0x4b0 crackme.01.32 | head -14
08048000 31ED xor ebp,ebp
08048002 5E pop esi
08048003 89E1 mov ecx,esp
08048005 83E4F0 and esp,byte -0x10
08048008 50 push eax
08048009 54 push esp
0804800A 52 push edx
0804800B 6800880408 push dword 0x8048800
08048010 6890870408 push dword 0x8048790
08048015 51 push ecx
08048016 56 push esi
08048017 68C8860408 push dword 0x80486c8
0804801C E89FFFFFFF call dword 0x8047fc0
08048021 F4 hlt
mitsurugi@mitsu:~/crackmes$
This looks like pretty usual stuff. The 0x080486c8 should be the address of the main function.
Before tracing main, let's find some offset of some strings to help:
at 0x86a : Please tell me my password:
at 0x888 : The password is correct! Congratulations!!!
at 0x8b5 : No! No! No! No! Try again.
Now we try to find those references in the main function, because we know that somewhere, it should asks for a password, so we should see those addresses pushed on the stack and the function printf or puts called.
mitsurugi@mitsu:~/crackmes$ ndisasm -au -o 0x080486c8 -e 0x6c8 crackme.01.32 | head -41
080486C8 55 push ebp
080486C9 89E5 mov ebp,esp
080486CB 83E4F0 and esp,byte -0x10
080486CE 83EC20 sub esp,byte +0x20
080486D1 A1A09B0408 mov eax,[0x8049ba0]
080486D6 8944240C mov [esp+0xc],eax
080486DA C74424081C000000 mov dword [esp+0x8],0x1c
080486E2 C744240401000000 mov dword [esp+0x4],0x1
080486EA C704246A880408 mov dword [esp],0x804886a
address of "Please tell me my password"
080486F1 E83AFDFFFF call dword 0x8048430
It should be puts function
080486F6 A1809B0408 mov eax,[0x8049b80]
080486FB 89442408 mov [esp+0x8],eax
080486FF C744240407000000 mov dword [esp+0x4],0x7
08048707 8D442418 lea eax,[esp+0x18]
0804870B 890424 mov [esp],eax
0804870E E8FDFCFFFF call dword 0x8048410
08048713 8D442418 lea eax,[esp+0x18]
08048717 890424 mov [esp],eax
0804871A E823FFFFFF call dword 0x8048642
0804871F C7442404609B0408 mov dword [esp+0x4],0x8049b60
08048727 8D442418 lea eax,[esp+0x18]
0804872B 890424 mov [esp],eax
0804872E E842FFFFFF call dword 0x8048675
08048733 85C0 test eax,eax
08048735 7527 jnz 0x804875e
goes to loose func
08048737 A1A09B0408 mov eax,[0x8049ba0]
0804873C 8944240C mov [esp+0xc],eax
08048740 C74424082C000000 mov dword [esp+0x8],0x2c
08048748 C744240401000000 mov dword [esp+0x4],0x1
08048750 C7042488880408 mov dword [esp],0x8048888
address of winning string
08048757 E8D4FCFFFF call dword 0x8048430
puts function again
0804875C EB25 jmp short 0x8048783
0804875E A1A09B0408 mov eax,[0x8049ba0]
08048763 8944240C mov [esp+0xc],eax
08048767 C74424081B000000 mov dword [esp+0x8],0x1b
0804876F C744240401000000 mov dword [esp+0x4],0x1
08048777 C70424B5880408 mov dword [esp],0x80488b5
address of loose message
0804877E E8ADFCFFFF call dword 0x8048430
and puts function again
08048783 B800000000 mov eax,0x0
08048788 C9 leave
08048789 C3 ret
mitsurugi@mitsu:~/crackmes$
Things becomes clearer now. Everything important is here:
080486EA C704246A880408 mov dword [esp],0x804886a | Please tell me my password
080486F1 E83AFDFFFF call dword 0x8048430 | should be puts function
080486F6 A1809B0408 mov eax,[0x8049b80]
080486FB 89442408 mov [esp+0x8],eax
080486FF C744240407000000 mov dword [esp+0x4],0x7
08048707 8D442418 lea eax,[esp+0x18]
0804870B 890424 mov [esp],eax
0804870E E8FDFCFFFF call dword 0x8048410
A function is called, it would be logical that gets is called to retrieve the password entered by the user.
08048713 8D442418 lea eax,[esp+0x18]
08048717 890424 mov [esp],eax
0804871A E823FFFFFF call dword 0x8048642
Now it's interesting, what this function does? Password validation? Password transformation? Other?
0804871F C7442404609B0408 mov dword [esp+0x4],0x8049b60
Another point of interest. What we have at this address?
08048727 8D442418 lea eax,[esp+0x18]
0804872B 890424 mov [esp],eax
0804872E E842FFFFFF call dword 0x8048675
Another function? Validation of password?
08048733 85C0 test eax,eax
Makes sense: if validation function return something different than 0, go loose.
08048735 7527 jnz 0x804875e
Yep, it goes to loose function
So, we have to check addresses:
-0x08048410 : is it gets?
-0x08048642 : what happens there?
-0x08049b60 : what is there?
-0x08048675 : is it a validation function? How does it works?
Let's go:
mitsurugi@mitsu:~/crackmes$ ndisasm -au -o 0x08048410 -e 0x410 crackme.01.32 | head -3
08048410 FF25309B0408 jmp dword [dword 0x8049b30]
08048416 6800000000 push dword 0x0
0804841B E9E0FFFFFF jmp dword 0x8048400
mitsurugi@mitsu:~/crackmes$
Ok, typically a GOT entry. I assume it's puts function.
next one
mitsurugi@mitsu:~/crackmes$ ndisasm -au -o 0x08048642 -e 0x642 crackme.01.32 | head -19
08048642 55 push ebp
08048643 89E5 mov ebp,esp
08048645 83EC10 sub esp,byte +0x10
08048648 C745FC00000000 mov dword [ebp-0x4],0x0
0804864F EB1C jmp short 0x804866d
08048651 8B55FC mov edx,[ebp-0x4]
08048654 8B4508 mov eax,[ebp+0x8]
08048657 01C2 add edx,eax
08048659 8B4DFC mov ecx,[ebp-0x4]
0804865C 8B4508 mov eax,[ebp+0x8]
0804865F 01C8 add eax,ecx
08048661 0FB600 movzx eax,byte [eax]
08048664 83F06C xor eax,byte +0x6c
Doing a simple XOR with value 0x6c
08048667 8802 mov [edx],al
08048669 8345FC01 add dword [ebp-0x4],byte +0x1
0804866D 837DFC05 cmp dword [ebp-0x4],byte +0x5
Looping 6 times.
08048671 7EDE jng 0x8048651
08048673 C9 leave
08048674 C3 ret
mitsurugi@mitsu:~/crackmes$
So this function just XOR with 0x6c the 6 first chars of the input.
Next one is 0x08049b60, we must take care of the address. It's 0x08049b60, so it's in the second load section. and the second load section is loaded with an offset of 0x0849000. It means that in the file, the 0x08049b60 are located at 0xb60. Let's see with hexdump what we have:
00000b60 1b 04 15 02 5c 18 00 00 47 43 43 3a 20 28 47 4e |....\...GCC: (GN|
00000b70 55 29 20 34 2e 38 2e 30 00 00 2e 73 68 73 74 72 |U) 4.8.0...shstr|6 bytes with a null at the end.
Now it's time for a big guess: the 0x08048675 is a simple validation function like strcmp.
If I resume how the binary works:
1/ It asks for a password
2/ It XORs the 6 first chars of the password with 0x6c
3/ It checks if the results is 0x1b 0x04 0x15 0X02 0x5c 0x18
(If it doesn't work, it means that we have to go deeper and disassemble 0x08048675 to learn what's going on there.)
We can easily calculate:
mitsurugi@mitsu:~/crackmes$ cat sol.py
#! /usr/bin/python
v=''
for p in '\x1b\x04\x15\x02\x5c\x18':
v += chr((ord(p)^0x6c))
print v
mitsurugi@mitsu:~/crackmes$ ./sol.py
whyn0t
mitsurugi@mitsu:~/crackmes$
This results let think that all assumptions were rights, we can easily verify:
mitsurugi@mitsu:~/crackmes$ ./crackme.01.32
Please tell me my password: whyn0t
The password is correct!
Congratulations!!!
mitsurugi@mitsu:~/crackmes$
Prenons l'exemple du binaire crackme.01.32 de crackmes.de. L'antidebug est simple à réaliser et très efficace comme nous allons voir: la partie section header du binaire est erronée. Un hexdump le montre bien:
mitsurugi@mitsu:~/crackmes$ hexdump -C crackme.01.32 | head -4
00000000 7f 45 4c 46 01 01 01 00 00 00 00 00 00 00 00 00 |.ELF............|
00000010 02 00 03 00 ff ff ff ff b0 84 04 08 34 00 00 00 |............4...|
00000020 ff ff ff ff 00 00 00 00 ff ff 20 00 08 00 ff ff |.......... .....|
00000030 ff ff ff ff 06 00 00 00 34 00 00 00 34 80 04 08 |........4...4...|
mitsurugi@mitsu:~/crackmes$
Les sections headers sont remplacées par des ff ff, voir le post précédent avec le grand tableau de l'elf101.
Et c'est suffisant pour mettre hors jeu les debuggers classiques sous linux que j'ai pu essayer:
mitsurugi@mitsu:~/crackmes$ objdump -d crackme.01.32
objdump: crackme.01.32: File format not recognized
mitsurugi@mitsu:~/crackmes$ gdb -q crackme.01.32
"/home/mitsurugi/crackmes/crackme.01.32": not in executable format: Format de fichier non reconnu
(gdb) q
mitsurugi@mitsu:~/crackmes$ readelf -s crackme.01.32
readelf: Error: Unable to seek to 0xffffffff for section headers
readelf: Error: Unable to seek to 0xffffffff for section headers
mitsurugi@mitsu:~/crackmes$
Les debuggers spécialisés chargent le binaire, mais semblent avoir des problèmes par la suite. edb a crashé assez vite, radare2 charge le binaire, mais ne fait aucun mapping par la suite: tous octets du binaires sont simplement plaqué linéairement à partir de l'adresse 0x08048000. J'ai ouvert un bug chez radare2 du coup.
En fait, seul IDA semble bien s'en sortir. Le post de blog http://em386.blogspot.de/2006/10/elf-no-section-header-no-problem.html qui date de 2006 arrive exactement à la même conclusion.
Je trouve quand même ça fou que tous les debuggers s’appuient sur des sections qui ne sont pas utilisés à l'exécution. Cela signifie qu'il doit être possible de duper ces debuggers/reversers en forgeant des en-têtes de sections bien choisis. Comme le dit le post de blog plus haut, gdb est un peu dédié à l'analyse de code dont on a les sources, donc c'est normal qu'il ne soit pas robuste aux binaires mal fichus, mais c'est quand même dommage qu'il ne sache pas s'en sortir sans section headers.
J'ai envoyé une solution à crackmes.de en anglais en donnant une méthode sans IDA pour trouver la solution. Ca m'a permis de bien disséquer de l'ELF, mais c'est relativement long.
Mitsurugi
Solution:
Launch the binary to begin:
mitsurugi@mitsu:~/crackmes$ ./crackme.01.32
Please tell me my password: password
No! No! No! No! Try again.
mitsurugi@mitsu:~/crackmes$ gdb -q ./crackme.01.32
"/home/mitsurugi/crackmes/./crackme.01.32": not in executable format: File format not recognized
gdb$ q
mitsurugi@mitsu:~/crackmes$
At a first glance we can see an antidebug trick by wiping the section header:
mitsurugi@mitsu:~/crackmes$ readelf -s crackme.01.32
readelf: Error: Unable to seek to 0xffffffff for section headers
readelf: Error: Unable to seek to 0xffffffff for section headers
mitsurugi@mitsu:~/crackmes$
objdump fails at analyzing it, and gdb too. If we strings the binary we can also see:
mitsurugi@mitsu:~/crackmes$ strings -n10 crackme.01.32
/lib/ld-linux.so.2
_IO_stdin_used
__libc_start_main
__gmon_start__
I'm sorry GDB! You are not allowed!
Tracing is not allowed... Bye
Please tell me my password:
The password is correct!
Congratulations!!!
No! No! No! No! Try again.
GCC: (GNU) 4.8.0
.note.ABI-tag
.note.gnu.build-id
.gnu.version
.gnu.version_r
.eh_frame_hdr
.init_array
.fini_array
mitsurugi@mitsu:~/crackmes$
The message about GDB let us assume that there are some other anti-gdb tricks, ptrace maybe:
mitsurugi@mitsu:~/crackmes$ strings crackme.01.32 | grep ptrace
ptrace
mitsurugi@mitsu:~/crackmes$
So it's time to disass it.We know that section headers are not usefull at all when running a binary, but program header are mandatory. Let's see what we have:
mitsurugi@mitsu:~/crackmes$ readelf -l crackme.01.32
readelf: Error: Unable to seek to 0xffffffff for section headers
readelf: Error: Unable to seek to 0xffffffff for section headers
Elf file type is EXEC (Executable file)
Entry point 0x80484b0
There are 8 program headers, starting at offset 52
Program Headers:
Type Offset VirtAddr PhysAddr FileSiz MemSiz Flg Align
PHDR 0x000034 0x08048034 0x08048034 0x00100 0x00100 R E 0x4
INTERP 0x000134 0x08048134 0x08048134 0x00013 0x00013 R 0x1
[Requesting program interpreter: /lib/ld-linux.so.2]
LOAD 0x000000 0x08048000 0x08048000 0x00a28 0x00a28 R E 0x1000
LOAD 0x000a28 0x08049a28 0x08049a28 0x00140 0x00180 RW 0x1000
DYNAMIC 0x000a38 0x08049a38 0x08049a38 0x000e8 0x000e8 RW 0x4
NOTE 0x000148 0x08048148 0x08048148 0x00044 0x00044 R 0x4
GNU_EH_FRAME 0x0008d4 0x080488d4 0x080488d4 0x00044 0x00044 R 0x4
GNU_STACK 0x000000 0x00000000 0x00000000 0x00000 0x00000 RW 0x4
mitsurugi@mitsu:~/crackmes$
So we can map some memory addresses with file offset.
First LOAD section:
From 0x08048000 to 0x08048a28, it's a linear mapping with the file from offset 0. But for bytes next, we have to make a gap of 0x1000.
Second LOAD section:
From 0x08049a28 to 0x08049a28+0x180 we have to map offset from files 0xa2 to 0xa28+0x140..
The DYNAMIC section, NOTE, GNU_EH_FRAME are not used for solving this crackme.
And now, with the help of ndisasm, we can solve the challenge by starting at entry point:
mitsurugi@mitsu:~/crackmes$ ndisasm -au -o 0x08048000 -e 0x4b0 crackme.01.32 | head -14
08048000 31ED xor ebp,ebp
08048002 5E pop esi
08048003 89E1 mov ecx,esp
08048005 83E4F0 and esp,byte -0x10
08048008 50 push eax
08048009 54 push esp
0804800A 52 push edx
0804800B 6800880408 push dword 0x8048800
08048010 6890870408 push dword 0x8048790
08048015 51 push ecx
08048016 56 push esi
08048017 68C8860408 push dword 0x80486c8
0804801C E89FFFFFFF call dword 0x8047fc0
08048021 F4 hlt
mitsurugi@mitsu:~/crackmes$
This looks like pretty usual stuff. The 0x080486c8 should be the address of the main function.
Before tracing main, let's find some offset of some strings to help:
at 0x86a : Please tell me my password:
at 0x888 : The password is correct! Congratulations!!!
at 0x8b5 : No! No! No! No! Try again.
Now we try to find those references in the main function, because we know that somewhere, it should asks for a password, so we should see those addresses pushed on the stack and the function printf or puts called.
mitsurugi@mitsu:~/crackmes$ ndisasm -au -o 0x080486c8 -e 0x6c8 crackme.01.32 | head -41
080486C8 55 push ebp
080486C9 89E5 mov ebp,esp
080486CB 83E4F0 and esp,byte -0x10
080486CE 83EC20 sub esp,byte +0x20
080486D1 A1A09B0408 mov eax,[0x8049ba0]
080486D6 8944240C mov [esp+0xc],eax
080486DA C74424081C000000 mov dword [esp+0x8],0x1c
080486E2 C744240401000000 mov dword [esp+0x4],0x1
080486EA C704246A880408 mov dword [esp],0x804886a
address of "Please tell me my password"
080486F1 E83AFDFFFF call dword 0x8048430
It should be puts function
080486F6 A1809B0408 mov eax,[0x8049b80]
080486FB 89442408 mov [esp+0x8],eax
080486FF C744240407000000 mov dword [esp+0x4],0x7
08048707 8D442418 lea eax,[esp+0x18]
0804870B 890424 mov [esp],eax
0804870E E8FDFCFFFF call dword 0x8048410
08048713 8D442418 lea eax,[esp+0x18]
08048717 890424 mov [esp],eax
0804871A E823FFFFFF call dword 0x8048642
0804871F C7442404609B0408 mov dword [esp+0x4],0x8049b60
08048727 8D442418 lea eax,[esp+0x18]
0804872B 890424 mov [esp],eax
0804872E E842FFFFFF call dword 0x8048675
08048733 85C0 test eax,eax
08048735 7527 jnz 0x804875e
goes to loose func
08048737 A1A09B0408 mov eax,[0x8049ba0]
0804873C 8944240C mov [esp+0xc],eax
08048740 C74424082C000000 mov dword [esp+0x8],0x2c
08048748 C744240401000000 mov dword [esp+0x4],0x1
08048750 C7042488880408 mov dword [esp],0x8048888
address of winning string
08048757 E8D4FCFFFF call dword 0x8048430
puts function again
0804875C EB25 jmp short 0x8048783
0804875E A1A09B0408 mov eax,[0x8049ba0]
08048763 8944240C mov [esp+0xc],eax
08048767 C74424081B000000 mov dword [esp+0x8],0x1b
0804876F C744240401000000 mov dword [esp+0x4],0x1
08048777 C70424B5880408 mov dword [esp],0x80488b5
address of loose message
0804877E E8ADFCFFFF call dword 0x8048430
and puts function again
08048783 B800000000 mov eax,0x0
08048788 C9 leave
08048789 C3 ret
mitsurugi@mitsu:~/crackmes$
Things becomes clearer now. Everything important is here:
080486EA C704246A880408 mov dword [esp],0x804886a | Please tell me my password
080486F1 E83AFDFFFF call dword 0x8048430 | should be puts function
080486F6 A1809B0408 mov eax,[0x8049b80]
080486FB 89442408 mov [esp+0x8],eax
080486FF C744240407000000 mov dword [esp+0x4],0x7
08048707 8D442418 lea eax,[esp+0x18]
0804870B 890424 mov [esp],eax
0804870E E8FDFCFFFF call dword 0x8048410
A function is called, it would be logical that gets is called to retrieve the password entered by the user.
08048713 8D442418 lea eax,[esp+0x18]
08048717 890424 mov [esp],eax
0804871A E823FFFFFF call dword 0x8048642
Now it's interesting, what this function does? Password validation? Password transformation? Other?
0804871F C7442404609B0408 mov dword [esp+0x4],0x8049b60
Another point of interest. What we have at this address?
08048727 8D442418 lea eax,[esp+0x18]
0804872B 890424 mov [esp],eax
0804872E E842FFFFFF call dword 0x8048675
Another function? Validation of password?
08048733 85C0 test eax,eax
Makes sense: if validation function return something different than 0, go loose.
08048735 7527 jnz 0x804875e
Yep, it goes to loose function
So, we have to check addresses:
-0x08048410 : is it gets?
-0x08048642 : what happens there?
-0x08049b60 : what is there?
-0x08048675 : is it a validation function? How does it works?
Let's go:
mitsurugi@mitsu:~/crackmes$ ndisasm -au -o 0x08048410 -e 0x410 crackme.01.32 | head -3
08048410 FF25309B0408 jmp dword [dword 0x8049b30]
08048416 6800000000 push dword 0x0
0804841B E9E0FFFFFF jmp dword 0x8048400
mitsurugi@mitsu:~/crackmes$
Ok, typically a GOT entry. I assume it's puts function.
next one
mitsurugi@mitsu:~/crackmes$ ndisasm -au -o 0x08048642 -e 0x642 crackme.01.32 | head -19
08048642 55 push ebp
08048643 89E5 mov ebp,esp
08048645 83EC10 sub esp,byte +0x10
08048648 C745FC00000000 mov dword [ebp-0x4],0x0
0804864F EB1C jmp short 0x804866d
08048651 8B55FC mov edx,[ebp-0x4]
08048654 8B4508 mov eax,[ebp+0x8]
08048657 01C2 add edx,eax
08048659 8B4DFC mov ecx,[ebp-0x4]
0804865C 8B4508 mov eax,[ebp+0x8]
0804865F 01C8 add eax,ecx
08048661 0FB600 movzx eax,byte [eax]
08048664 83F06C xor eax,byte +0x6c
Doing a simple XOR with value 0x6c
08048667 8802 mov [edx],al
08048669 8345FC01 add dword [ebp-0x4],byte +0x1
0804866D 837DFC05 cmp dword [ebp-0x4],byte +0x5
Looping 6 times.
08048671 7EDE jng 0x8048651
08048673 C9 leave
08048674 C3 ret
mitsurugi@mitsu:~/crackmes$
So this function just XOR with 0x6c the 6 first chars of the input.
Next one is 0x08049b60, we must take care of the address. It's 0x08049b60, so it's in the second load section. and the second load section is loaded with an offset of 0x0849000. It means that in the file, the 0x08049b60 are located at 0xb60. Let's see with hexdump what we have:
00000b60 1b 04 15 02 5c 18 00 00 47 43 43 3a 20 28 47 4e |....\...GCC: (GN|
00000b70 55 29 20 34 2e 38 2e 30 00 00 2e 73 68 73 74 72 |U) 4.8.0...shstr|6 bytes with a null at the end.
Now it's time for a big guess: the 0x08048675 is a simple validation function like strcmp.
If I resume how the binary works:
1/ It asks for a password
2/ It XORs the 6 first chars of the password with 0x6c
3/ It checks if the results is 0x1b 0x04 0x15 0X02 0x5c 0x18
(If it doesn't work, it means that we have to go deeper and disassemble 0x08048675 to learn what's going on there.)
We can easily calculate:
mitsurugi@mitsu:~/crackmes$ cat sol.py
#! /usr/bin/python
v=''
for p in '\x1b\x04\x15\x02\x5c\x18':
v += chr((ord(p)^0x6c))
print v
mitsurugi@mitsu:~/crackmes$ ./sol.py
whyn0t
mitsurugi@mitsu:~/crackmes$
This results let think that all assumptions were rights, we can easily verify:
mitsurugi@mitsu:~/crackmes$ ./crackme.01.32
Please tell me my password: whyn0t
The password is correct!
Congratulations!!!
mitsurugi@mitsu:~/crackmes$
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